## Geometric Quantization (Part 6)

Now let’s do some more interesting examples of geometric quantization using the functor described in Part 4. Let’s look at the spin-j particle with j > 1/2.

To be specific, let’s consider the spin-3/2 particle. There’s nothing special about the number 3 here: everything I’ll say can be generalized. But the number 3 will give me a nice excuse to show you a picture of a curve called the ‘twisted cubic’.

We can build a spin-3/2 particle from three spin-1/2 particles, all having angular momenta pointing in the same direction. Classically this procedure amounts to tripling a vector, but quantum-mechanically it’s related to cubing. This is a bit mysterious to me, but let me explain.

Classically, if we take a vector of length 1/2 in $\mathbb{R}^3$ and triple it we get a vector of length 3/2. This gives a map from classical states of the spin-1/2 particle to classical states of the spin-3/2 particle. In other words: a map from the sphere of radius 1/2 to the sphere of radius 3/2. Simple! But this is not a symplectic map, since as we saw last time, we are giving the latter sphere a symplectic structure that’s 3 times as big. So, it’s not a valid classical process. Still, it’s a perfectly fine way to get our hands on lots of states of the classical spin-3/2 particle! All of them, in fact.

Quantum-mechanically the state space for the spin-1/2 particle is $\mathbb{C}^2.$ We can think of a guy in here as a linear functional on the dual ${\mathbb{C}^2}^\ast.$ We can cube this and get a homogeneous polynomial of degree 3 on ${\mathbb{C}^2}^\ast.$ The space of such polynomials is called $S^3(\mathbb{C}^2),$ and this is the space of states of the quantum spin-3/2 particle. So, we get a map

$\text{cubing} \colon \mathbb{C}^2 \to S^3(\mathbb{C}^2)$

Simple! This describes the process of ‘triplicating’ the state of spin-1/2 particle and getting a spin-3/2 particle. But this is not a linear map, so it’s not a valid quantum process. As the saying goes, “you can’t clone a quantum”. Still, it’s a perfectly fine way to get our hands on lots of states of the quantum spin-3/2 particle! But not all of them, as we’ll soon see.

Geometric quantization should reconcile and combine classical and quantum mechanics. How can we do this here?

It’s pretty simple. We projectivize the map

$\text{cubing} \colon \mathbb{C}^2 \to S^3(\mathbb{C}^2)$

Cubing sends any line through the origin in $\mathbb{C}^2$ into a unique line through the origin in $S^3(\mathbb{C}^2).$ So, we get a map from $\mathbb{C}\mathrm{P}^1$ to the projective space of $S^3(\mathbb{C}^2).$ And we define the image of this map to be the space of classical states of the spin-3/2 particle!

Let’s see what it looks like. I’m thinking of an element $(a,b) \in \mathbb{C}^2$ as a linear functional

$f(x,y) = ax + by$

If we cube it we get this homogeneous polynomial of degree 3:

$f(x,y)^3 = a^3 \, x^3 + 3a^2b \, x^2y + 3ab^2 \, xy^2 + b^3 \, y^3$

If we take $x^3, x^2y, xy^2$ and $y^3$ as our basis of the homogeneous polynomials of degree 3, we can identify $S^3(\mathbb{C}^2)$ with $\mathbb{C}^4.$ So we get

$\begin{array}{clll} \text{cubing} \colon & \mathbb{C}^2 &\to& \mathbb{C}^4 \\ & (a,b) & \mapsto & (a^3, 3a^2b, 3ab^2, b^3) \end{array}$

If we projectivize this map we get a map between projective varieties, which I’ll call

$P(\text{cubing}) \colon \mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^3$

The image of this map is the space of states of the classical spin 3/2-particle! It’s a copy of $\mathbb{C}\mathrm{P}^1$ sitting inside projective 3-space. But because cubing a nonlinear map, this copy will be twisted: not a ‘line’ but a ‘curve’. People call it the twisted cubic.

Just for kicks, let’s take a closer look at it. Let’s use homogeneous coordinates and write $[a,b]$ for the point in $\mathbb{C}\mathrm{P}^1$ corresponding to a nonzero vector $(a,b) \in \mathbb{C}^2.$ Similarly, any point in $\mathbb{C}\mathrm{P}^3$ can be written as a nonzero 4-tuple of complex numbers with a bracket around it. We get

$\begin{array}{cccl} P(\text{cubing}) \colon & \mathbb{C}\mathrm{P}^1 &\to& \mathbb{C}\mathrm{P}^3 \\ & [a,b] & \mapsto & [a^3, 3a^2b, 3ab^2, b^3] \end{array}$

so the twisted cubic is

$\{ [a^3, 3a^2b, 3ab^2, b^3] : \, (0,0) \ne (a,b) \in \mathbb{C}^2 \} \subset \mathbb{C}\mathrm{P}^3$

This is hard to visualize, so we can work with a copy of $\mathbb{C}^3$ that’s dense inside $\mathbb{C}\mathrm{P}^3,$ namely the set where $a = 1.$ The portion of the twisted cubic sitting in this set is

$\{ (b, 3b^2, b^3) : \, b \in \mathbb{C} \} \subset \mathbb{C}^3$

This is still hard to visualize, so we can restrict to the reals and think about the curve

$\{ (b, 3b^2, b^3) : \, b \in \mathbb{R} \} \subset \mathbb{R}^3$

This is also called the twisted cubic! People often rescale the $y$ axis to get rid of the number 3 here.

This version of the twisted cubic is still a bit hard to visualize, but it’s the intersection of two very nice surfaces: the surface $y = x^2$ and the surface $z = x^3.$ So, the twisted cubic is the black curve:

in this nice picture uploaded to ResearchGate by Alexander M. Kasprzyk.

Okay, back to serious business! Let’s call the twisted cubic $C_3 \subseteq \mathbb{C}\mathrm{P}^3.$ It’s exactly the sort of thing we can geometrically quantize using our functor

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

The reason is that it’s a projective variety and it’s linearly normal. Moreover when we quantize $C_3$ we get $\mathbb{C}^4,$ which is just what we want for the spin-3/2 particle!

$\texttt{Q}(C_3) = \mathbb{C}^4$

Why? Remember, $\texttt{Q}(C_3)$ is defined to be the smallest linear subspace $V \subseteq \mathbb{C}^4$ such that $C_3 \subseteq P(V).$ But $C_3$ twists around so much that the smallest $V$ that works is all of $\mathbb{C}^4.$

Now let’s take stock of where we are and draw some general conclusions from what we’ve seen in this example. We now have a firm grip on the space of quantum states of the spin-3/2 particle:

$S^3(\mathbb{C}^2) \cong \mathbb{C}^4$

and also the space of classical states of the spin-3/2 particle, the twisted cubic:

$C_3 = \{ [a^3, 3a^2b, 3ab^2, b^3] : \, (0,0) \ne (a,b) \in \mathbb{C}^2 \}$

Now for something cool: the latter sits inside the projectivization of the former! This is obvious, but it has a very nice physical meaning. While I’ve been calling $\mathbb{C}^4$ the space of quantum states of the spin-3/2 particle, it is very reasonable to argue that quantum states are actually points of its projectivization, $\mathbb{C}\mathrm{P}^3.$ I will skip the argument, which is old, famous and convincing:

• Wikipedia, Projective Hilbert space.

What matters for us here is that classical states of the spin-3/2 particle give some of these quantum states—far from all, but some of the nicest ones! They are the states obtained by ‘cubing’ a state of a spin-1/2 particle. In other words, they are the states where we can think of our spin-3/2 particle as made of three spin-1/2 particles with their spins perfectly aligned.

It’s nice to say this with a bit more physics jargon. These quantum states coming from classical ones are called ‘coherent states’ . A general quantum state of the spin-3/2 particle is a ‘quantum superposition’ of these coherent states. By this, I mean that the smallest linear subspace $V \subseteq \mathbb{C}^4$ for which $P(V)$ contains the twisted cubic is all of $\mathbb{C}^4.$

And while we’ve seen this in a particular example, it’s a completely general feature of our setup! Remember, projectivization

$\texttt{P} \colon \texttt{Quant} \to \texttt{Class}$

is the left adjoint of quantization

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

This means that for any $M \in \texttt{Class}, V \in \texttt{Quant}$ we have

$\texttt{Q}(M) \subseteq V \quad \iff \quad M \subseteq \texttt{P}(V)$

We get something interesting if we take $V = \texttt{Q}(M)$ here. Since $\texttt{Q}(M) \subseteq \texttt{Q}(M),$ we get

$M \subseteq \texttt{P} (\texttt{Q}(M))$

Category theorists call this inclusion of $M$ in $\texttt{P} (\texttt{Q}(M))$ the ‘unit’ of our pair of adjoint functors. It says how classical states sit inside the projectivization of $\texttt{Q}(M).$ If we call points of $\texttt{P}(\texttt{Q}(M))$ quantum states, those in $M$ are called the coherent states.

Moreover every quantum state is a ‘quantum superposition’ of coherent states! In other words, the smallest linear subspace $V \subseteq \texttt{Q}(M)$ for which $M$ sits inside $\texttt{P}(V)$ is all of $\texttt{Q}(M).$

Why is this true? It’s just the definition of $\texttt{Q}(M)!$

So, I hope you see how much physics is packed into these adjoint functors $\texttt{Q}$ and $\texttt{P}.$ I’ll do more examples next time. In the meantime, you can read more about this approach to the spin-3/2 particle here:

• Dorje C. Brody and Lane P. Hughston, Geometric quantum mechanics.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.

### 11 Responses to Geometric Quantization (Part 6)

1. Layra Idarani says:

You keep saying that the space of quantum states of a spin 3/2 particle is $\mathbb{C}^3$ even though a few times you say $\mathbb{C}^4$ on occasion. Should it be $\mathbb{C}^4$ throughout?

• John Baez says:

I meant to always write $\mathbb{C}^4$ since 4 is the dimension of the space of homogeneous cubic polynomials in 2 variables. I kept slipping and typing $\mathbb{C}^3$ since I had $\mathbb{C}\mathrm{P}^3$, cubics, and spin 3/2 on my mind. I thought I caught all instances of $\mathbb{C}^3$ and eliminated them, but apparently not. I’ll try again.

• John Baez says:

Okay, done. There should be just two instances of $\mathbb{C}^3$ left, but I really mean those!

2. Layra Idarani says:

After the first $\mathbb{C}^3$, there is a $\mathbb{C}\textbf{P}^1$ that should be a $\mathbb{C}\textbf{P}^3$.

3. strangeset says:

Hi John, visited your ucr blog, amazing stuff. I am going to be a regular reader and yes I would like to help save the planet, do tell me how can I contribute?

I teach, math, physics a bit of coding and have been dabbling in data analysis and network biology last 2-3 years.

• John Baez says:

My new theory is that you (like everyone) have a great potential to do something to save the planet, not by dropping what you are currently doing or adding extra volunteer activities to your existing job, but taking your current profession and adapting it to focus more energy on this grand task. The point is that saving the planet requires many different people to do many different things, and it will work best if we each do something we’re already good at—and even paid to do!

Let’s see if we can test this theory. You say you teach. Teachers are in a great position to learn about the Anthropocene and what to do about it, and then multiply their knowledge by passing some of it on to their students. We can incorporate interesting examples drawn from these issues in almost any course. The point here is not to propagandize but to give students the mental tools they’ll need to do well and help the planet do well.

Of course there are also other things you could do outside work, but if you teach you’re in a great position to do something really big through that.

The Azimuth Library has a lot of useful information about environmental issues, and you could learn from it and also add to it.

4. Last time I looked at an example, where I built the spin-3/2 particle by cloning the spin-1/2 particle.

In brief, it went like this. The space of classical states of the spin-1/2 particle is the Riemann sphere, $\mathbb{C}\mathrm{P}^1.$ This just happens to also be the space of quantum states of the spin-1/2 particle, since it’s the projectivization of $\mathbb{C}^2.$ To get the 3/2 particle we look at the map

$\text{cubing} \colon \mathbb{C}^2 \to S^3(\mathbb{C}^2)$

Now, you’ll have noted that the numbers 2 and 3 show up in what I just said. But there’s nothing special about these numbers! They could be arbitrary natural numbers… well, > 1 if we don’t enjoy thinking about degenerate cases.

Here’s how the generalization works [….]

5. Is there a particular motivation behind the picture of Cayley’s nodal cubic in this post? (Aside from it being a cute picture in the realm of algebraic geometry). I actually know a rather precise way in which said surface shows up in QM, but not in the realm of geometric quantization.

• John Baez says:

Each post in this series has a cute picture of an algebraic variety. You can click on it to learn more about it. The picture in this post is not Cayley’s nodal cubic; it’s the Endrass octic.

I don’t know what happens when you geometrically quantize the varieties in these pictures—but that’s something I hope to discover! Most of them aren’t smooth, so they go beyond the basic formalism I’m sketching here, which is designed to minimized technicalities. But we can go ahead and geometrically quantize non-smooth algebraic varieties, and this should be very interesting!

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